Minus Minus makes a Plus

“Minus minus makes a plus.” You have all heard a version of this at some point in your life. Your teacher was trying to explain integers to you and this phrase was supposed to help you understand. The phrase stems from this problem:

4 – (-3) =

This question is often difficult for students. If the question was simply 4 – 3, then the answer would be 1. However, the extra negative sign confounds many students. A quick teacher response is to rattle off the phrase minus minus makes a plus. Then the student can transform the problem and solve it with ease:

4 – (-3)

4 + 3 = 7

Yesterday, I was trying to teach this concept to my wife and she kept asking why. She understood how to do the questions but she wanted a deeper explanation as to why the two minuses made a plus. Unfortunately, even though I have thought long and hard about integers, I have been unable to explain why the trick works. That is, until yesterday.

After going back and forth with her for a while, I suddenly had an idea! I will illustrate it in picture form. 4 can be represented as 4 individual +1s.

Picture 1

How can we subtract -3 from the above picture? There are no -1s to take away. Here was my epiphany. We need to create an environment in which this subtraction can take place. Consider the new picture below:

picture 2

The picture still has the original 4 in the middle. However, there are now a bunch of +1 -1 pairs all around the 4. Since +1 – 1 = 0, these pairs do not change the answer and the picture still represents 4. Now the beauty of this approach.

We need to calculate 4 – (-3). That means we need to subtract -3. Hence, we need to subtract 3 -1s. Like so:

Picture 3

We have subtracted (or removed) 3 -1s. Count the total. The answer is clearly +7. This technique and be used with positive or negative numbers using addition or subtraction.

Here is another example:

-5 – (-7)

First the initial setup:

Picture 4

Then the subtraction:

Picture 5

The result is +2. I hope the above explanation gives you a deeper understanding of why subtracting a negative is equivalent to adding a positive. I know it did for me.

Leave a comment

Filed under Uncategorized

Percentage Activity

I was at teacher’s conference (WestCAST 2015) the other week and I heard an interesting idea for how to teach percentages. After tweaking the idea for a little while, I think I am ready to share it.

A percent is a part of a hundred. The word comes from the Latin adverbial phrase per centum meaning by the hundred.1 Suppose we have 100 people in a room. If 40 are men and 60 are women we say 40 percent are men and 60 percent are women. We often use the symbol “%” instead of the word “percent.”

Consider the following picture of Mario:


Let’s investigate different sections of Mario and determine the percentage of each colour. To do this, we will use a sheet of paper with a 10 by 10 hole cut in it. Thus, we will only ever see 100 squares at a time. Suppose I place the paper and this is what I see.

Mario covered

Let’s count the number of times each colour appears:

Tan: 7

Red: 15

Blue: 39

White: 39

First, we notice that 7 + 15 + 39 + 39 = 100 which is good because there are supposed to be 100 squares! Second, it is as easy as adding a percent symbol to turn these numbers into percentages.

Tan: 7%

Red: 15%

Blue: 39%

White: 39%

I really like this activity for a number of reasons. First, it clearly communicates that percentages are proportions of a hundred. Second, each student can place the 10 by 10 hole on a different part of Mario and get a different answer. Third, if the students are interested in the concept, they can design their own pictures in Excel to create percentage activities.

As a follow up activity, we could ask the question, what is the percentage of white squares in the entire picture of Mario? How can we calculate percent when we have more than 100 squares?

1 http://dictionary.reference.com/browse/percent

1 Comment

Filed under Uncategorized

Complex Roots Part 2

In the last post, we had poor Carlos trying to catch the bus. To understand the third situation we need a bit of knowledge of square roots. You see, when we take square roots, we usually only allow positive numbers. For example:

\sqrt{9} =3

But \sqrt{-9} is not usually defined. Now, if you had to try to define \sqrt{-9} , you would want it to be something like 3, since it looks kind of like it should be 3. However, you also want to differentiate it from the regular 3. So why not just add a symbol and call it:


Before you declare this approach crazy, consider this. What is 10-3 ?

10-3 = 7 of course.

What if we reverse the order. What is 3-10 ?

-7 of course.

Normally in subtraction you have a bigger number and you take away a smaller number. When the situation changes we simply introduce a new symbol and invent a new word, “negative”.

I hope that justifies my methods. If not, just humour me for now.

Returning to Carlos, when I use the formula for the first situation I get the values 2 and 5. As you can see from the graph, these are points of when the blue line (Carlos) red line (the bus) intersect.

2014-11-04 10.59.25

When I use the formula for the second situation, I get the value 3.16. Again, this is the point where Carlos catches up with the bus (where the two lines intersect).

2014-11-04 11.04.22

When I use the formula for the third situation I get an interesting answer. I get:

2.5 + i1.9

What does this solution mean? It can’t mean that Carlos catches the bus, because the lines do not intersect. However, Carlos does get close to the bus. The proper interpretation is as follows.

The first number, 2.5, means that after 2.5 seconds Carlos will be as close as possible to the bus. Before 2.5 seconds, he was gaining on the bus and after 2.5 seconds the bus gets further and further away. The second number, 1.9, means that at 2.5 seconds, Carlos is 1.9 meters away from the bus. This is the closest that Carlos will get to the bus. If Carlos had a friend on the bus with a really long pole, maybe he could grab on to the bus. But otherwise, he will miss it.

2014-11-04 11.06.01

Even though Carlos does not make his bus, the math can still tell us interesting information about the situation. This is preferable to the usual “no solution” we are taught in school.

What I have used are the numbers more formally know as the complex numbers. While some people might say they are imaginary numbers, I say, that real or not, they are incredibly useful.

Leave a comment

Filed under Uncategorized

Complex roots Part 1

Carlos has not had the best morning. First, he was out of milk for his cereal. Then he splattered mustard his shirt while frantically making his lunch for the day. By the time he left his house, he realized he might even miss the bus.

As Carlos rounded the last corner, he saw his bus. He instantly started running after it at full speed, 10m/s. At the same time, the bus revved up its engines and started accelerating at a rate of 5m/s2. The bus already had a 10 meter head start on Carlos. Will Carlos catch the bus?

Problems like these are typical in a pre-calculus class. The usual solution is to model the distance between the bus and Carlos as a function of time. Then use some method to solve the equation. Rather than delving too deep into the mathematical process, I want to highlight a few situations graphically with you.

Suppose Carlos can run really fast. It is possible that he could catch up to the bus and pass the bus. However, eventually the bus would get up to speed and would pass Carlos. This situation can be modelled by the following graph. The red line is the position of the bus and the blue line is the position of Carlos.


2014-11-04 10.59.25

Maybe Carlos isn’t the best runner. Maybe he just barely catches up with the bus. The graph below shows this possibility.

2014-11-04 11.04.22

Finally, maybe the bus is too fast, or had too much of a head start, or Carlos had a stomach cramp. Whatever the case, Carlos misses his bus. This situation is graphed below.

2014-11-04 11.06.01

The first and second possibilities are usually the ones that are emphasized in class. You can use a formula (the quadratic formula) to find out what happens to Carlos, how long he has to run to catch the bus, where exactly he catches it, and so on. The third situation is simply left as “no solution.” But that is not entirely true. Although Carlos does not catch up with the bus, he does get closer to it. In fact, there is a point when he gets really close, but then the bus zooms away. If you work through the formula, you will see that that you actually find out how close Carlos got to the bus and at what time. The only problem is that we have to cheat a bit. Stay tuned for part 2.

1 Comment

Filed under Uncategorized

Lattice Polgyons Part 2

In the previous post, I introduced the idea of a lattice polygon and a method for calculating the area. I want to discuss a very powerful theorem.

Before we can state the theorem, we need to identify two types of points.

  • A boundary point (B) is a point that lies on one of the lines of the lattice polygon
  • An interior point (I) is a point that is contained inside the lattice polygon

In the lattice polygon below the boundary points are in blue and the interior points are in red.

polygon 1

This lattice polygon has B = 11 (blue points) and I = 6 (red points)

The theorem, formulated in 1899 by Georg Pick, states the following:

Area = \frac{1}{2}*B+I-1

Therefore, the theorem states that our previous lattice polygon has an area of:

Area = \frac{1}{2}*11+6-1 = 10.5

You can use the method from the previous post and verify that the area is indeed 10.5. This theorem is cool because you can take any huge and complicated lattice polygon and find the area with a super simple formula. We have no need to resort to cutting up the shape into smaller triangles and rectangles. For example, find the area of the following lattice polygon:

polygon 2

You should find B = 25, I = 10, and area = 21.5. Pretty cool.

Leave a comment

Filed under Uncategorized

Lattice Polygons Part 1

A lattice polygon:

polygon 1

It has straight sides, and each point lies on a lattice (grid). Lattice polygons are really fun because we can practice area calculations without getting messy numbers. For example: consider the following lattice polygon:

polygon 2

Since it is a rectangle, we can use the formula arearectangle = base * height which in this case is 2*3 = 6. We could also have counted the 6 squares inside the shape. Ok, that was too easy; we need to move on to a more complicated shape. How about the following lattice polygon:

polygon 3

If you use the square counting method, then the area is clearly 8. How could we apply our formula to get the same result? The shape is not a rectangle.

If you take a shape and cut it into pieces, those pieces will have the same area as the original shape. Therefore, if we are clever enough with our cuts, we can turn the shape into a bunch of smaller rectangles.

polygon 4

Now we have:

  • a rectangle with base 1 and height 1 (area 1)
  • a rectangle with a base of 3 and a height of 1 (area 3)
  • a rectangle with a base of 2 and a height of 2 (area 4)

Putting that all together, we have the following area: 1 + 3 + 4 = 8

Time to ramp it up a notch. We need one more formula and things will get crazy. The area of a triangle is areatriangle = 1/2 * base * height

Consider the original lattice polygon. Here is how we could split it up:

Polygon 5

We have the following:

  • a triangle of base 3 and height 1 (area 1.5)
  • a triangle of base 1 and height 1 (area 0.5)
  • a triangle of base 2 and height 2 (area 2)

Putting that all together, we have the following area: 1.5 + 0.5 + 2 = 4

One last, massive, lattice polygon:

Polygon 6

Here is one way to split up the giant shape:

Polygon 7

If you do all of the relevant calculations, you will find that the area is 18. Stay tuned for the next post.

Leave a comment

Filed under Uncategorized


I remember a time long ago, elementary school, when the school would hold a contest. There were different varieties of this contest but the basic version went as follows.

Guess the number of M&Ms contained in the jar. The person closest wins the jar!


The question is simple but to get anything better than a random guess, it’s best if we apply some mathematics.

First, we need to find the volume of container. We will assume that the jar is cylinder. Maybe it is 20 cm tall and 10 cm in radius. Therefore, the volume is calculated using the formula:

V = \pi * r^2 * h = 3.14 * 10^2 * 20 = 6280 cm^3

Now we need to find the volume of each M&M. A quick Google search gives us a volume of 0.636 cm3 for each M&M. Now we divide these two quantities to determine how many M&Ms fit in the container, 6280 / 0.636 = 9874.

However, this answer is too large. The above equation is assuming that the container is completely filled with M&Ms. If you look closely, you will see that there are little gaps between the pieces that are filled with air and not candy. To account for this, we need to take a quick detour.

Consider the following problem, how many of these circles can you fit into this square without overlapping?


picture 1

Here is one attempt where I get 6 circles:

picture 2

Here is one where I can get 7 circles:

picture 3

The best packing I can get is 8 circles:

picture 4

In this configuration, the circles take up 73% of the total area. We can use the same concept with the M&Ms. For packing circular objects into a container, the percentage is 64%. This means that we have a total of:

9874 * 64% = 9874 * 0.64 = 6319

There you have it, there should be 6319 M&Ms. Next time you are guessing at the jar, use a little math.


1 Comment

Filed under Uncategorized